Question

如果我错了，请纠正我，但是......

给定任意源平面的法线，以及应用所需旋转后平面的法线：

Vector3F sourceNormal = (x, y, z).normalize()
Vector3F desiredNormal = (0, 0, 1).normalize()

1）我们可以通过两个法线的交叉积来找到“旋转轴”

Vector3F rotationAxis = Vector3F.cross(sourceNormal, desiredNormal).normalize()

2）我们可以通过两个法线的点积的弧余弦找到“旋转角”。

// Thanks nico - it was there in my project source, but it was omitted here.
float theta = Math.acos(Vector3F.dot(sourceNormal, desiredNormal))

3）我们可以将旋转应用于一组点，以便将源平面定向到我们想要的平面。

float[] rotationMatrix = new float[16];

// X component
rotationMatrix[5] = rotationMatrix[10] = (float)Math.cos(theta);
rotationMatrix[9] = (float)Math.sin(theta);
rotationMatrix[6] = -rotationMatrix[9];

// Y component
rotationMatrix[0] = rotationMatrix[10] = (float)Math.cos(theta);
rotationMatrix[2] = (float)Math.sin(theta);
rotationMatrix[8] = -rotationMatrix[2];

// Z component
rotationMatrix[0] = rotationMatrix[5] = (float)Math.cos(theta);
rotationMatrix[1] = (float)Math.sin(theta);
rotationMatrix[4] = -rotationMatrix[1];

for(Point3F point : polygon)
{
    float x = pt.getX();
    float y = pt.getY();
    float z = pt.getZ();
    float[] xs = new float[3];
    float[] ys = new float[3];
    float[] zs = new float[3];
    for(int j = 0; j < 3; ++j)
    {
    xs[j] = rotationMatrix[j] * x;
    ys[j] = rotationMatrix[j + 4] * y;
    zs[j] = rotationMatrix[j + 8] * z;
    }
    x = 0; y = 0; z = 0;
    for(int j = 0; j < 3; ++j)
    {
    x += xs[j];
    y += ys[j];
    z += zs[j];
    }
    pt.set(x, y, z);
}

我的输出不正确。

点数：

(-56.00, 72.01, 48.02)
(-48.00, 72.01, 48.02)
(-48.00, 86.01, 24.02)
(-56.00, 86.01, 24.02)

Out Points：

(-124.960010, -88.105451, 24.185812)
(-107.108590, -88.105451, 24.185812)
(-107.108590, -105.237051, 12.0929052)
(-124.960010, -105.237051, 12.0929052)

如果我不得不猜测，我会说我正在不正确地应用旋转点...也许我已经错误地解释了本文中的旋转矩阵（http://en.wikipedia.org/wiki/Rotation_matrix）？

感谢您的任何意见。

...假设这是设置旋转矩阵的正确方法，输出仍然不正确：

Vector3F axis = Vector3F.cross(sourceNormal, desiredNormal).normalize();
float angle = (float) Math.acos(p.normal.dot(new Vector3F(0, 0, 1)));
float s = (float)Math.sin(angle);
float c = (float)Math.cos(angle);
float x = axis.getX(), y = axis.getY(), z = axis.getZ();
float[] matrix = new float[16];
matrix[0] = x * x * (1 - c) + c;
matrix[1] = x * y * (1 - c) - (z * s);
matrix[2] = x * z * (1 - c) + (y * s);

matrix[4] = y * x * (1 - c) + (z * s);
matrix[5] = y * y * (1 - c) + c;
matrix[6] = y * z * (1 - c) - (x * s);

matrix[8] = x * z * (1 - c) - (y * s);
matrix[9] = y * z * (1 - c) + (x * s);
matrix[10] = z * z * (1 - c) + c;


float nx = x * matrix[0] + y * matrix[1] + z * matrix[2];
float ny = x * matrix[4] + y * matrix[5] + z * matrix[6];
float nz = x * matrix[8] + y * matrix[9] + z * matrix[10];

In: (-56.00, 56.01, -16.02)
In: (-48.00, 56.01, -16.02)
In: (-48.00, 72.01, -8.02)
In: (-56.00, 72.01, -8.02)
Out: (-51.270340, 46.887921, -25.5108342)
Out: (-43.9460070, 46.887921, -25.5108342)
Out: (-43.9460070, 62.798761, -21.5554182)
Out: (-51.270340, 62.798761, -21.5554182)

Answer 1

您的旋转矩阵错误。

旋转需要根据您链接到的维基百科文章的第10部分中的公式进行组合（“轴和角度的旋转矩阵”）。

此矩阵的简化版也可以在glRotate函数的手册页中找到。

注意：除非您反复重复轮换，否则无需再次acos(dotp)取cos和sin。将点积直接用作cos(theta)，并使用sin(theta)^2 = 1 - cos(theta)^2关系计算sin(theta)。

Answer 2

最终，有很多问题。 Alnitak解决了绝大多数问题，我刚刚找到了最后一个。

简而言之，以下是错误的：

1）矩阵创建

2）矩阵乘法

3）浮点截断

即使它可能是一个hacky解决方案，它对我的目的也没关系。工作片段如下。

Vector3F axis = new Vector3F(p.normal);
// Seems to work when adding precision...but NOT with (0, 0, 1)
Vector3F target = new Vector3F(0.000000000000000001f, 0.000000000000001f, 1.0f);
axis.cross(target);
axis.normalize();
float angle = (float) Math.acos(p.normal.dot(new Vector3F(0, 0, 1)));
float s = (float) Math.sin(angle);
float c = (float) Math.cos(angle);
float x = axis.getX(), y = axis.getY(), z = axis.getZ();
float[] matrix = new float[16];
matrix[0] = x * x * (1 - c) + c;
matrix[1] = x * y * (1 - c) - (z * s);
matrix[2] = x * z * (1 - c) + (y * s);
matrix[4] = y * x * (1 - c) + (z * s);
matrix[5] = y * y * (1 - c) + c;
matrix[6] = y * z * (1 - c) - (x * s);
matrix[8] = x * z * (1 - c) - (y * s);
matrix[9] = y * z * (1 - c) + (x * s);
matrix[10] = z * z * (1 - c) + c;

float nx = x * matrix[0] + y * matrix[1] + z * matrix[2];
float ny = x * matrix[4] + y * matrix[5] + z * matrix[6];
float nz = x * matrix[8] + y * matrix[9] + z * matrix[10];

// triangulate (now 2D) polygon
// Rotate the points back
angle = -angle;
s = (float) Math.sin(angle);
c = (float) Math.cos(angle);
matrix[0] = x * x * (1 - c) + c;
matrix[1] = x * y * (1 - c) - (z * s);
matrix[2] = x * z * (1 - c) + (y * s);
matrix[4] = y * x * (1 - c) + (z * s);
matrix[5] = y * y * (1 - c) + c;
matrix[6] = y * z * (1 - c) - (x * s);
matrix[8] = x * z * (1 - c) - (y * s);
matrix[9] = y * z * (1 - c) + (x * s);
matrix[10] = z * z * (1 - c) + c;
nx = x * matrix[0] + y * matrix[1] + z * matrix[2];
ny = x * matrix[4] + y * matrix[5] + z * matrix[6];
nz = x * matrix[8] + y * matrix[9] + z * matrix[10];
// All's well that ends well

旋转任意平面为z轴对齐

2 个答案: